Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
To solve the given system of linear equations:
[tex]\[ 4x - 2y = 8 \quad \text{and} \quad y = \frac{3}{2}x - 2 \][/tex]
we can follow the steps provided.
### Step 1: Plot the [tex]\( x \)[/tex]-intercept of the first equation.
To find the [tex]\( x \)[/tex]-intercept of the equation [tex]\( 4x - 2y = 8 \)[/tex], set [tex]\( y = 0 \)[/tex]:
[tex]\[ 4x - 2(0) = 8 \\ 4x = 8 \\ x = 2 \][/tex]
So the [tex]\( x \)[/tex]-intercept is at the point [tex]\( (2, 0) \)[/tex].
### Step 2: Plot the [tex]\( y \)[/tex]-intercept of the first equation.
To find the [tex]\( y \)[/tex]-intercept of the first equation, set [tex]\( x = 0 \)[/tex]:
[tex]\[ 4(0) - 2y = 8 \\ -2y = 8 \\ y = -4 \][/tex]
So the [tex]\( y \)[/tex]-intercept is at the point [tex]\( (0, -4) \)[/tex].
### Step 3: Plot the [tex]\( y \)[/tex]-intercept of the second equation.
For the second equation [tex]\( y = \frac{3}{2}x - 2 \)[/tex], set [tex]\( x = 0 \)[/tex]:
[tex]\[ y = \frac{3}{2}(0) - 2 \\ y = -2 \][/tex]
So the [tex]\( y \)[/tex]-intercept is at the point [tex]\( (0, -2) \)[/tex].
### Step 4: Use the slope of the second equation to plot one more point.
The slope of the second equation, [tex]\( y = \frac{3}{2}x - 2 \)[/tex], is [tex]\( \frac{3}{2} \)[/tex]. This means for every increase in [tex]\( x \)[/tex] by 1, [tex]\( y \)[/tex] increases by [tex]\( \frac{3}{2} \)[/tex].
Starting from the [tex]\( y \)[/tex]-intercept [tex]\( (0, -2) \)[/tex], if [tex]\( x \)[/tex] increases by 1:
[tex]\[ x = 1 \\ y = \frac{3}{2}(1) - 2 \\ y = \frac{3}{2} - 2 \\ y = -0.5 \][/tex]
So, another point is [tex]\( (1, -0.5) \)[/tex].
### Conclusion:
From the above steps, we have identified the following points:
- [tex]\( (0, -4) \)[/tex] from the first equation's [tex]\( y \)[/tex]-intercept.
- [tex]\( (0, -2) \)[/tex] from the second equation's [tex]\( y \)[/tex]-intercept.
- [tex]\( (2, 0) \)[/tex] from the first equation's [tex]\( x \)[/tex]-intercept.
- [tex]\( (1, -0.5) \)[/tex] from the slope of the second equation.
Putting these points into the table, we get:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & -4 \\ 0 & -2 \\ 2 & 0 \\ 1 & -0.5 \\ \hline \end{array} \][/tex]
So, we have plotted the intercepts and a few other key points which would help in graphically solving or verifying the system of equations.
[tex]\[ 4x - 2y = 8 \quad \text{and} \quad y = \frac{3}{2}x - 2 \][/tex]
we can follow the steps provided.
### Step 1: Plot the [tex]\( x \)[/tex]-intercept of the first equation.
To find the [tex]\( x \)[/tex]-intercept of the equation [tex]\( 4x - 2y = 8 \)[/tex], set [tex]\( y = 0 \)[/tex]:
[tex]\[ 4x - 2(0) = 8 \\ 4x = 8 \\ x = 2 \][/tex]
So the [tex]\( x \)[/tex]-intercept is at the point [tex]\( (2, 0) \)[/tex].
### Step 2: Plot the [tex]\( y \)[/tex]-intercept of the first equation.
To find the [tex]\( y \)[/tex]-intercept of the first equation, set [tex]\( x = 0 \)[/tex]:
[tex]\[ 4(0) - 2y = 8 \\ -2y = 8 \\ y = -4 \][/tex]
So the [tex]\( y \)[/tex]-intercept is at the point [tex]\( (0, -4) \)[/tex].
### Step 3: Plot the [tex]\( y \)[/tex]-intercept of the second equation.
For the second equation [tex]\( y = \frac{3}{2}x - 2 \)[/tex], set [tex]\( x = 0 \)[/tex]:
[tex]\[ y = \frac{3}{2}(0) - 2 \\ y = -2 \][/tex]
So the [tex]\( y \)[/tex]-intercept is at the point [tex]\( (0, -2) \)[/tex].
### Step 4: Use the slope of the second equation to plot one more point.
The slope of the second equation, [tex]\( y = \frac{3}{2}x - 2 \)[/tex], is [tex]\( \frac{3}{2} \)[/tex]. This means for every increase in [tex]\( x \)[/tex] by 1, [tex]\( y \)[/tex] increases by [tex]\( \frac{3}{2} \)[/tex].
Starting from the [tex]\( y \)[/tex]-intercept [tex]\( (0, -2) \)[/tex], if [tex]\( x \)[/tex] increases by 1:
[tex]\[ x = 1 \\ y = \frac{3}{2}(1) - 2 \\ y = \frac{3}{2} - 2 \\ y = -0.5 \][/tex]
So, another point is [tex]\( (1, -0.5) \)[/tex].
### Conclusion:
From the above steps, we have identified the following points:
- [tex]\( (0, -4) \)[/tex] from the first equation's [tex]\( y \)[/tex]-intercept.
- [tex]\( (0, -2) \)[/tex] from the second equation's [tex]\( y \)[/tex]-intercept.
- [tex]\( (2, 0) \)[/tex] from the first equation's [tex]\( x \)[/tex]-intercept.
- [tex]\( (1, -0.5) \)[/tex] from the slope of the second equation.
Putting these points into the table, we get:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & -4 \\ 0 & -2 \\ 2 & 0 \\ 1 & -0.5 \\ \hline \end{array} \][/tex]
So, we have plotted the intercepts and a few other key points which would help in graphically solving or verifying the system of equations.
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
